Week 4
Introduction
Today’s exercise exercise will focus on different techniques of clustering and classification. I will use data on housing in areas of Boston and mostly focus on the crime rate in the city. The data can be accessed through the R package MASS. The data contains area-level information on the characteristics of homes (size, value etc.), the demographic composition of the area as well as several variables related to environmental and infrastructural factors. More information on the data is available here and in the original study by Harrison & Rubinfeld (1978).
BTW, I decided to hide my code chunks by default in this course diary as they make reading a bit tedious. You should be able to get the codes by clicking the Code button next to results.
Overview of the dataset
library(MASS)
library(tidyverse)
library(GGally)
library(corrplot)
data(Boston)
dim(Boston)## [1] 506 14str(Boston)## 'data.frame': 506 obs. of 14 variables:
## $ crim : num 0.00632 0.02731 0.02729 0.03237 0.06905 ...
## $ zn : num 18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
## $ indus : num 2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
## $ chas : int 0 0 0 0 0 0 0 0 0 0 ...
## $ nox : num 0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
## $ rm : num 6.58 6.42 7.18 7 7.15 ...
## $ age : num 65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
## $ dis : num 4.09 4.97 4.97 6.06 6.06 ...
## $ rad : int 1 2 2 3 3 3 5 5 5 5 ...
## $ tax : num 296 242 242 222 222 222 311 311 311 311 ...
## $ ptratio: num 15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
## $ black : num 397 397 393 395 397 ...
## $ lstat : num 4.98 9.14 4.03 2.94 5.33 ...
## $ medv : num 24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08205 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00hist(Boston$crim)
The data includes 14 variables from some 500 regions in Boston. All of them are numeric.My main varibale of interest, crime rate, seems highly skewed, with most of the areas having low rates of crime and a few expressing higher rates.
p.values.mat <-cor.mtest(Boston,
conf.level = .95)
cor.mat <- cor(Boston)
corrplot.mixed(cor.mat,
lower.col='black',
upper='color',
tl.col='black',
tl.cex=0.5,
number.cex=0.5,
p.mat=p.values.mat$p,
sig.level=0.05)
Correlation plot
For a graphical overview of data, I am using the correlation plot to make the plot to some extent readable (compared to pairs or ggpairs). In the plot, I have crossed out all the correlations not significant at 95% confidence level. It seems that the variable chas is not significantly correlated with most of the variables (probably as it is binary). Anyhow, the variable indicates whether the area is bounded by river Charles, which makes little sense to me.
Most of the other variables are statistically correlated, and there seems to be relatively strong correlations, for instance property tax rate (tax) and access to highways (rad) have a correlation of 0.91, age (age) of houses and distance from city centre (dis) have a correlation of -0.75, and nitrous ocygen emissions and proportion of non-retail businesses (=industry) have a correlation of 0.76. These are rather intuitive. Most of the correlations seem moderate, between 0.3 and 0.6.The highest correlations between crime rate occur for variables access to radial highways and property tax rate.
Linear Discriminant analysis
As a next part of the assignment, I am running a Linear Discriminant Analysis (LDA). LDA allows for classifying observations to pre-known categories, based on linear associations between variables in the data.There are two assumptions in the model, the variables are normally distributed and has the same variance. Thus, the process starts by scaling the data.The effects of scaling can be seen on the following summary: all the variables are centered around their mean, which is now 0.
#Scale boston data
boston_scaled <- as.data.frame(scale(Boston))
summary(boston_scaled)## crim zn indus chas
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563 Min. :-0.2723
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668 1st Qu.:-0.2723
## Median :-0.390280 Median :-0.48724 Median :-0.2109 Median :-0.2723
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150 3rd Qu.:-0.2723
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202 Max. : 3.6648
## nox rm age dis
## Min. :-1.4644 Min. :-3.8764 Min. :-2.3331 Min. :-1.2658
## 1st Qu.:-0.9121 1st Qu.:-0.5681 1st Qu.:-0.8366 1st Qu.:-0.8049
## Median :-0.1441 Median :-0.1084 Median : 0.3171 Median :-0.2790
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.5981 3rd Qu.: 0.4823 3rd Qu.: 0.9059 3rd Qu.: 0.6617
## Max. : 2.7296 Max. : 3.5515 Max. : 1.1164 Max. : 3.9566
## rad tax ptratio black
## Min. :-0.9819 Min. :-1.3127 Min. :-2.7047 Min. :-3.9033
## 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876 1st Qu.: 0.2049
## Median :-0.5225 Median :-0.4642 Median : 0.2746 Median : 0.3808
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058 3rd Qu.: 0.4332
## Max. : 1.6596 Max. : 1.7964 Max. : 1.6372 Max. : 0.4406
## lstat medv
## Min. :-1.5296 Min. :-1.9063
## 1st Qu.:-0.7986 1st Qu.:-0.5989
## Median :-0.1811 Median :-0.1449
## Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.6024 3rd Qu.: 0.2683
## Max. : 3.5453 Max. : 2.9865Next, I will first calculate target classes for the LDA model. I am using the crime rate in the areas and dividing it into quartiles. Then, I divide the data into train and test sets by randomly sampling 80% of the areas (train). The rest of the areas are included in the test set.Lastly, I fit the LDA model on the training data and test the model by conducting predictions with the test set
#Save categories of crime rate
bins <- quantile(boston_scaled$crim)
#Create new crime variable
crime <- cut(boston_scaled$crim,
breaks=bins,
include.lowest=T,
label=c(
"low",
"med_low",
"med_high",
"high"))
boston_scaled$crim <- NULL
boston_scaled$crime <- crime
#Divide data into test and training sets
n <- nrow(boston_scaled)
#Randomly sample 80% of the original rows
#These are used for training
ind <- sample(n, size=n*0.8)
#Train set
train <- boston_scaled[ind,]
#Test set
test <- boston_scaled[-ind,]
#Correct classes in the test set
correct <- test$crime
#Drop crime from test
test <- dplyr::select(test, -crime)
#LDA model
lda.fit <-
lda(crime~., data=train)
# the function for lda biplot arrows
#(Stolen from Datacamp)
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "red", tex = 0.75, choices = c(1,2)){
heads <- coef(x)
arrows(x0 = 0, y0 = 0,
x1 = myscale * heads[,choices[1]],
y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
text(myscale * heads[,choices], labels = row.names(heads),
cex = tex, col=color, pos=3)
}
# target classes as numeric
classes <- as.numeric(train$crime)
# predict classes with test data
lda.pred <- predict(lda.fit,
newdata = test)Let’s have a look at the results.
# plot the lda results
plot(lda.fit, dimen = 2,
col=classes,
pch=classes)
lda.arrows(lda.fit, myscale = 2)
#summary of the model
lda.fit## Call:
## lda(crime ~ ., data = train)
##
## Prior probabilities of groups:
## low med_low med_high high
## 0.2623762 0.2351485 0.2549505 0.2475248
##
## Group means:
## zn indus chas nox rm age
## low 0.9940435 -0.9058593 -0.12375925 -0.8848690 0.4066166 -0.9336949
## med_low -0.1212048 -0.2774201 -0.02367011 -0.5508559 -0.1389731 -0.3058610
## med_high -0.3773415 0.1643633 0.10991367 0.3686017 0.0224771 0.3700800
## high -0.4872402 1.0171519 -0.03610305 1.0720351 -0.4635588 0.8311561
## dis rad tax ptratio black lstat
## low 0.9032850 -0.6882537 -0.7286418 -0.49845065 0.3750736 -0.74466322
## med_low 0.3697372 -0.5442448 -0.4755179 0.03366711 0.3554805 -0.13446476
## med_high -0.3354462 -0.4221329 -0.3167420 -0.22476388 0.1064134 0.03879449
## high -0.8691570 1.6377820 1.5138081 0.78037363 -0.8901512 0.93618249
## medv
## low 0.5272604
## med_low 0.0045591
## med_high 0.1128689
## high -0.7102025
##
## Coefficients of linear discriminants:
## LD1 LD2 LD3
## zn 0.182270586 0.7979106628 -0.904467987
## indus -0.060470552 -0.1963362060 0.292297312
## chas -0.005582401 0.0135462199 0.112735109
## nox 0.457189696 -0.6738283206 -1.400675765
## rm 0.023973158 -0.0698027106 -0.128343291
## age 0.276848494 -0.3713227787 -0.003276884
## dis -0.139602247 -0.3327638885 0.323917998
## rad 3.068199288 1.0896143626 -0.001208730
## tax 0.054263045 -0.1950561946 0.570889726
## ptratio 0.178630494 -0.0392931271 -0.140908671
## black -0.144020238 -0.0009138888 0.169074168
## lstat 0.165635367 -0.2314381116 0.360724123
## medv 0.056470967 -0.4092119322 -0.089237640
##
## Proportion of trace:
## LD1 LD2 LD3
## 0.9493 0.0390 0.0116# cross tabulate
#the correct classes vs. predictions
table(correct = correct,
predicted = lda.pred$class)## predicted
## correct low med_low med_high high
## low 9 10 2 0
## med_low 8 13 10 0
## med_high 0 3 19 1
## high 0 0 0 27table(correct = correct,
predicted = lda.pred$class) %>%
prop.table(2) %>% round(digits=2)## predicted
## correct low med_low med_high high
## low 0.53 0.38 0.06 0.00
## med_low 0.47 0.50 0.32 0.00
## med_high 0.00 0.12 0.61 0.04
## high 0.00 0.00 0.00 0.96The most relevant information is included in the scatterplot, aka biplot, of the first two linear discriminants. The plot shows that the grouping with these LDAs of high crime rate areas was relatively successful, whereas the other groups tend to have more overlap. The arrows in the plot show the importance of each variable, and to which direction the are working. It seems that the access to radiaal highways sparates relatively well the high crime rate group. The summary of the model shows the same: mean of rad is high in the high crime rate group, and lower in others.The rad variable has also a large coefficient from the LD1. These findings indicate that crime occurs in places where there are easy to access highways, or, rather, in places where people come and go (probably city centres and some sort of knots in the public transportation system).
K-means clustering
#Reload boston
data(Boston)
boston_scaled <-
as.data.frame(scale(Boston))
#Calculate distances between observations
#I use Euclidean for no specific reason
#except that the km algorithm uses it
#by default
distances <- dist(boston_scaled)
summary(distances)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.1343 3.4625 4.8241 4.9111 6.1863 14.3970#Identify correct number of clusters
#Use the WCSS for this purpose
k_max <- 15 #Arbitrary number
twcss <-
sapply(1:k_max,
function(k){
kmeans(
boston_scaled,k)$tot.withinss})
plot(x=1:k_max,y=twcss,type='l')
#2 seems appropriate
#Run k-means algorithm
km <- kmeans(boston_scaled,centers=2)
#Plot the data set in three parts
pairs(boston_scaled[
c(1,5,6,7,8,12,13,14)],
col=km$cluster)
K-means + LDA
km2 <- kmeans(boston_scaled,centers=3)
boston_scaled$km_clust <- km2$cluster
#LDA model (rename to avoid confusion
#in the next step)
lda.fit2 <-
lda(km_clust~., data=boston_scaled)
lda.fit2## Call:
## lda(km_clust ~ ., data = boston_scaled)
##
## Prior probabilities of groups:
## 1 2 3
## 0.3992095 0.2806324 0.3201581
##
## Group means:
## crim zn indus chas nox rm
## 1 -0.3549295 -0.4039269 0.009294842 0.11748284 0.01531993 -0.2547135
## 2 0.9693718 -0.4872402 1.074440092 -0.02279455 1.04197430 -0.4146077
## 3 -0.4071299 0.9307491 -0.953383032 -0.12651054 -0.93243813 0.6810272
## age dis rad tax ptratio black
## 1 0.3096462 -0.2267757 -0.5759279 -0.4964651 -0.09219308 0.2473725
## 2 0.7666895 -0.8346743 1.5010821 1.4852884 0.73584205 -0.7605477
## 3 -1.0581385 1.0143978 -0.5976310 -0.6828704 -0.53004055 0.3582008
## lstat medv
## 1 0.09168925 -0.1052456
## 2 0.85963373 -0.6874933
## 3 -0.86783467 0.7338497
##
## Coefficients of linear discriminants:
## LD1 LD2
## crim -0.03654114 0.20373943
## zn 0.08346821 0.34784463
## indus 0.32262409 -0.12105014
## chas 0.04761479 -0.13327215
## nox 0.13026254 0.15610984
## rm -0.13267423 0.44058946
## age 0.11936644 -0.84880847
## dis -0.23454618 0.58819732
## rad 1.96894437 0.57933028
## tax 1.10861600 0.53984421
## ptratio 0.13087741 -0.02004405
## black -0.15432491 -0.06106305
## lstat 0.14002173 0.14786473
## medv -0.02559139 0.37307811
##
## Proportion of trace:
## LD1 LD2
## 0.8999 0.1001classes <-
as.numeric(boston_scaled$km_clust)
# plot the lda results
plot(lda.fit2, dimen = 2,
col=classes,
pch=classes)
lda.arrows(lda.fit2, myscale = 2)
model_predictors <-
dplyr::select(train, -crime)
# check the dimensions
dim(model_predictors)## [1] 404 13dim(lda.fit$scaling)## [1] 13 3# matrix multiplication
matrix_product <-
as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)
library(plotly)
plot_ly(x = matrix_product$LD1,
y = matrix_product$LD2,
z = matrix_product$LD3,
type= 'scatter3d',
mode='markers',
color=train$crime)#let's fit still another k-means
#For some reason, the code does not work
#with the original train data
#However, this just loads the boston
#and limirts the same areas so no
#errors here, I guess
data(Boston)
boston_scaled <- as.data.frame(scale(Boston))
train <- boston_scaled[ind,]
km3 <- kmeans(train, centers=2)
plot_ly(x = matrix_product$LD1,
y = matrix_product$LD2,
z = matrix_product$LD3,
type= 'scatter3d',
mode='markers',
color=km3$cluster)References
Harrison, D. & Rubinfeld, D.L. Hedonic housing prices and the demand for clean air. 1978. Journal of Environmental Economics and Management 5(1), 81-102.